# Can Maxwell's equations be derived from Coulomb's Law and Special Relativity?

As an exercise I sat down and derived the magnetic field produced by moving charges for a few contrived situations. I started out with Coulomb's Law and Special Relativity. For example, I derived the magnetic field produced by a current $I$ in an infinite wire. It's a relativistic effect; in the frame of a test charge, the electron density increases or decreases relative to the proton density in the wire due to relativistic length contraction, depending on the test charge's movement. The net effect is a frame-dependent Coulomb field whose effect on a test charge is exactly equivalent to that of a magnetic field according to the Biot–Savart Law.

My question is: Can Maxwell's equations be derived using only Coulomb's Law and Special Relativity?

If so, and the $B$-field is in all cases a purely relativistic effect, then Maxwell's equations can be re-written without reference to a $B$-field. Does this still leave room for magnetic monopoles?

• I have a vague recollection of, when I was in high school, finding a book that did undergrad E&M by assuming SR is correct from the beginning and doing something like this. I don't recall the title, though (or if it was any good), but if you want to see this worked out in detail you might try to look for it? – Mr X Jan 22 '11 at 18:05
• @Jeremy -- The book you're thinking of is probably Electricity and Magnetism by E. Purcell (part of the Berkeley Physics series). A very good book, by the way. – Ted Bunn Jan 22 '11 at 18:32
• Yes! I believe it is the one I was thinking of. – Mr X Jan 22 '11 at 18:33
• You also need the assumption that charge is a scalar and same-charges repel. Then the derivation is contained in Purcell's EM book. – Ron Maimon Jun 15 '12 at 2:06
• Yes, it is possible and it is carried out in great detail by Haskell in this document: richardhaskell.com/files/… See also the subsequent very illuminating discussion about the problems when applying Maxwell's equations to accelerated source charges. – exchange Nov 24 '17 at 19:54

Maxwell's equations do follow from the laws of electricity combined with the principles of special relativity. But this fact does not imply that the magnetic field at a given point is less real than the electric field. Quite on the contrary, relativity implies that these two fields have to be equally real.

When the principles of special relativity are imposed, the electric field $\vec{E}$ has to be incorporated into an object that transforms in a well-defined way under the Lorentz transformations - i.e. when the velocity of the observer is changed. Because there exists no "scalar electric force", and for other technical reasons I don't want to explain, $\vec{E}$ can't be a part of a 4-vector in the spacetime, $V_{\mu}$.

Instead, it must be the components $F_{0i}$ of an antisymmetric tensor with two indices, $$F_{\mu\nu}=-F_{\nu\mu}$$ Such objects, generally known as tensors, know how to behave under the Lorentz transformations - when the space and time are rotated into each other as relativity makes mandatory.

The indices $\mu,\nu$ take values $0,1,2,3$ i.e. $t,x,y,z$. Because of the antisymmetry above, there are 6 inequivalent components of the tensor - the values of $\mu\nu$ can be $$01,02,03;23,31,12.$$ The first three combinations correspond to the three components of the electric field $\vec{E}$ while the last three combinations carry the information about the magnetic field $\vec{B}$.

When I was 10, I also thought that the magnetic field could have been just some artifact of the electric field but it can't be so. Instead, the electric and magnetic fields at each point are completely independent of each other. Nevertheless, the Lorentz symmetry can transform them into each other and both of them are needed for their friend to be able to transform into something in a different inertial system, so that the symmetry under the change of the inertial system isn't lost.

If you only start with the $E_z$ electric field, the component $F_{03}$ is nonzero. However, when you boost the system in the $x$-direction, you mix the time coordinate $0$ with the spatial $x$-coordinate $1$. Consequently, a part of the $F_{03}$ field is transformed into the component $F_{13}$ which is interpreted as the magnetic field $B_y$, up to a sign.

Alternatively, one may describe the electricity by the electric potential $\phi$. However, the energy density from the charge density $\rho=j_0$ has to be a tensor with two time-like indices, $T_{00}$, so $\phi$ itself must carry a time-like index, too. It must be that $\phi=A_0$ for some 4-vector $A$. This whole 4-vector must exist by relativity, including the spatial components $\vec{A}$, and a new field $\vec{B}$ may be calculated as the curl of $\vec{A}$ while $\vec{E}=-\nabla\phi-\partial \vec{A}/\partial t$.

You apparently wanted to prove the absence of the magnetic monopoles by proving the absence of the magnetic field itself. Well, apologies for having interrupted your research plan: it can't work. Magnets are damn real. And if you're interested, the existence of magnetic monopoles is inevitable in any consistent theory of quantum gravity. In particular, two poles of a dumbbell-shaped magnet may collapse into a pair of black holes which will inevitably possess the (opposite) magnetic monopole charges. The lightest possible (Planck mass) black holes with magnetic monopole charges will be "proofs of concept" heavy elementary particles with magnetic charges - however, lighter particles with the same charges may sometimes exist, too.

• Your previous answer was only unclear in that it does not address why I am able (in a few contrived examples, admittedly) to solve for the equations of motion without reference to a B-field. All I need to do is show how the E-field transforms under Lorentz boosts, and I can do that without introducing a B-field. Did I not do those examples correctly, or are they lucky exceptions because they are contrived? – user1247 Jan 22 '11 at 18:20
• I read your answer thoroughly, and you are still not answering my questions. When I do my examples, nowhere must I postulate a "new field". I merely start with Coulomb's law and SR, and do the math, and the math shows that a particle experiences forces that can be effectively described by a "new field". This is analogous to the Coriolis force. Does gravity plus a rotating reference frame imply a new "Coriolis field"? Of course not, but it can be effectively described by one. – user1247 Jan 22 '11 at 18:44
• Lubos answer is very good and very precise. I fully subscribe. I am only a bit puzzled by the last paragraph of the answer, where he says that the existence of magnetic monopoles is inevitable in any consistent theory of quantum gravity. The argument given is that two poles of a dumbbell-shaped magnet may collapse into a pair of black holes which will "inevitably possess the (opposite) magnetic monopole charges". If I break a magnet, I do not get two monopoles, of course. I get two magnets. What would prevent the two black holes to do exactly the same, and behave like two magnets, without – Carlo Rovelli Jan 27 '11 at 8:07
• ...any monopole around? (that's the rest of Dr. Rovelli's comment, which was cut off by the system) – David Z Jan 27 '11 at 8:09
• Let me add some references. Look e.g. at arxiv.org/abs/hep-th/9404076 which refers to the paper of Affleck-Manton who generalized the Schwinger effect to magnetic fields. Just like the Schwinger electric field will produce electron-positron pairs, magnetic field will produce monopole-antimonopole pairs. The conditions allowing this production are inevitable in QG where the monopoles may be represented by monopole-charged black holes. – Luboš Motl Jan 27 '11 at 10:50

Lubos Motl's answer is very good, but I think it's worth saying one or two additional things.

You can regard magnetism as simply a byproduct of electricity, in the following sense: if you assume that Coulomb's Law is correct, and that special relativity is correct, and that charge is a Lorentz scalar (so that charge and current density form a 4-vector), then you can derive all of Maxwell's equations. (Actually, you probably also need to assume the theory is linear as well, now that I think about it.) The undergraduate-level textbook by Purcell works this out very explicitly in a nice, pleasing way, and it's also in more advanced textbooks.

Some books gloss over the need to postulate that charge is a scalar. At least one textbook -- I don't remember which -- does emphasize it, and makes a convincing case that it's worth paying attention to. One way to see that it's not a trivial condition to impose is to consider the analogy with gravity -- that is, substitute mass for charge and gravity for electric field, and try to run the same argument. (Assume weak fields so that everything can be treated as linear if you like.) There are "gravitomagnetic" effects, but they're not related to regular gravity in the same way as the magnetic field is related to the electric field -- i.e., the gravitational analogues of Maxwell's equations look different from the regular Maxwell equations). One reason is the sign differences, of course -- like charges repel in one case and attract in the other. But a bigger reason is that the source of gravity is not a scalar: its density doesn't form part of a 4-vector, but rather of a rank-2 tensor.

But on a more philosophical (or perhaps semantic) level, I wouldn't jump from this fact to the conclusion that magnetism is "merely" a byproduct of electricity. At the very least, such language doesn't appear to be useful in understanding the theory or in using it! For instance, understanding how an electromagnetic wave can propagate from a distant galaxy to your eye is much easier and more natural if you look at it from the "usual" point of view.

• Thanks Ted. So, if, as you say, you can derive all of Maxwell's equations as a byproduct of electricity, it seems to follow trivially that one can write Maxwell's equations without reference to a B-field. (Just as I can write, as in the exercise I described, the forces due to a current I without reference to a B-field). This is my question that Lubos seems to refuse to want to address. I understand that this doesn't change the physics, and may not be the most parsimonious way of expressing electromagnetism -- I'm just interested if it can and has been done. – user1247 Jan 23 '11 at 1:39
• Yes, it can be done. With sufficient effort, you can go further and express all of electricity and magnetism without reference to either an E or B field -- just as a very strange and complicated force law between charges, in which the force on each charge depends on the properties of the other charge at the retarded time. Griffiths's textbook writes the force law out explicitly in one of the later chapters. You give up a lot by doing this -- the biggest thing that comes to mind is that I have no idea how you'd even try to talk about energy-momentum conservation in this language. – Ted Bunn Jan 23 '11 at 16:03
• @user1247 Keep in mind that parsimony is extremely important. There's an infinite number of theories that explain the same phenomenon just as well and make the same predictions. We're trying to cut everything down to the fundamentals. You could replace all the quantum fields with just one rather complicated field that explained everything exactly as well, but despite replacing multiple fields with just one, it would be far more complex. And we expect that the fundamental laws governing the universe are as simple as possible (indeed, "fundamental complexity" sounds like an oxymoron :D ). – Luaan Aug 8 '18 at 10:50

Not a direct answer to your question but still a surprising derivation of Maxwell's equations:

Feynman's proof of the Maxwell equations (FJ Dyson - Phys. Rev. A, 1989) shows, that it is possible to derive Maxwell's equations from Newton's second law of motion and commutation relations (under non-relativistic limits).

• Cool! A copy is available here: scribd.com/doc/168392117/… . May be illegal, depending on the laws on copyright and fair use in a given country. – user4552 Sep 15 '13 at 20:47
• Have you read the comments that follow? Not all of the equations can be derived from them. Though it's nifty, it isn't complete. – Arturo don Juan Nov 2 '15 at 0:26

Yes, you can make it, but you also need to use a superposition principle.

1. You determine that Couloms's law, $$\mathbf F = \frac{qQ\mathbf r}{|\mathbf r |^{3}},$$ is a boundary case of the relativistic force, which acts on the charge q by the field of a Q-charge.
2. Using Lorentz transformation for the force and for the radius-vector, $$\mathbf F = \mathbf F' + \gamma \mathbf u \frac{(\mathbf F' \cdot \mathbf v')}{c^{2}} + \Gamma \mathbf u \frac{(\mathbf u \cdot \mathbf F')}{c^{2}},$$ $$\mathbf r' = \mathbf r + \Gamma \mathbf u \frac{(\mathbf u \cdot \mathbf r)}{c^{2}} - \gamma \mathbf u t = \mathbf r + \Gamma \mathbf u \frac{(\mathbf u \cdot \mathbf r)}{c^{2}} (t = 0),$$ where u is the speed of inertial system, v is the charge speed, you can assume, that relative to the other inertial system with relative speed u the force looks as $$\mathbf F = q\mathbf E + \frac{q}{c}[\mathbf v \times \mathbf B],$$ where $$\mathbf E = \frac{\gamma Q \mathbf r}{(r^{2} + \frac{\gamma^{2}}{c^{2}}(\mathbf r \cdot \mathbf u)^{2})^{\frac{3}{2}}}, \quad \mathbf B = \frac{1}{c}[\mathbf u \times \mathbf E].$$ Of course, magnetic field is a relativistiс kinematic effect, but a procedure described above are the relativistic kinematiс transformation of Coulomb's law. So some people made a mistake by giving negative answer.
3. After that, using primary theoremes of vector analisys and regularization procedure, you can "take" rot and div of the E and B expressions above. After that you can earn Maxwell's equations. You must use superposition principle, when you move from a field of one charge to multi-charge continuously distribution.
• +1 for both yours and WIMPs answer: Lubos and all the rest are of course quite right, but it of course depends on what you mean by derive. With the postulates of the OP treated as a formal axiom system: of course you can't; from a physicist's standpoint, where you assume other "reasonable" stuff like linearity, of course you can. – Selene Routley Oct 4 '14 at 3:41
• Note that this isn't actually a proof of Maxwell's equations. Special relativity only applies to charges which have (and always had) constant velocity, however Maxwell's equations can describe more than that. For example in 3. you prove that applying div to E leads to one of the Maxwell's equations. However it's not clear that this still works if the charge(s) were accelerated at any time and retarded fields come into play. So this is an additional assumtion of your prove, that it still holds for accelerated charges. – stonar96 Mar 28 at 12:13

I know that Purcell and others have used Lorentz symmetry as a pedagogical device to motivate the introduction of magnetic fields, but I do not recall ever having seen an axiomatic derivation of Maxwell's equations. It might be an interesting exercise to see precisely what assumptions beyond Lorentz symmetry and Coulomb's Law are necessary to reconstruct Maxwell's equations.

B fields are not fictitious fields

If you know the electric and magnetic fields in one inertial frame, you can determine the electric and magnetic fields in any other frame via Lorentz transformation. If the magnetic field happens to vanish in a given inertial frame, you could think of magnetic effects in other frames as fictitious. However, it is not always possible to find a frame in which the magnetic fields vanish. The fastest way to see this is to note that E^2 - B^2 c^2 is a Lorentz invariant quantity (see Wikipedia). If we find that B^2 > E^2/c^2 at a given spacetime point in a given inertial frame, it follows that B^2 > 0 at that point in all inertial frames. In fact, you could begin in a frame where the electric field vanishes but the magnetic field does not; the electric fields observed in other frames could then be considered fictitious.

In general, neither the electric field nor the magnetic field can be made to vanish under a Lorentz boost. To see this quickly, note that the dot product of the E field vector with the B field vector at a given spacetime point is a Lorentz invariant quantity (see Wikipedia). If this dot product is nonzero at a given spacetime point in a given inertial frame, the electric and magnetic field vectors will both be nonzero at that spacetime point in all inertial frames.

As Einstein pointed out, you can understand the motion of a charged particle by referring to the electric field in the rest frame of that particle. However, if you have multiple particles with different velocities, you need to keep track of the electric field in the instantaneous rest frame of each particle. Since Lorentz boosts mix the E field with the B field, the only way to keep track of the E field in the rest frame of each of your particles in terms of local quantities in one inertial frame is by reference to the E field and the B field.

Locality

Even if it is possible, it is not clear to me that it would be desirable to use Coulomb's law as an axiom in electromagnetic theory. Maxwell's equations explain the motion of particles by referring to local degrees of freedom, the fields. Coulomb's law, on the other hand, is a form of action-at-a-distance, and is manifestly non-local.

It is certainly possible to rewrite both the E and B fields in terms of integrals over charge density and current density (I can't post another link, so google "Jefimenko's equations"), and then to use these expressions to interpret electromagnetic forces as a form of retarded action-at-a-distance. However, to obtain these expressions requires assumptions about the boundary conditions on the E and B fields. We can always obtain another valid solution of Maxwell's equations by simply changing the boundary conditions on the fields, which demonstrates that the fields have independent existence, and are not mere book-keeping variables to simplify a more fundamental non-local interaction.

Monopoles

As usually written, Maxwell's equations do not contain terms corresponding to magnetic charge, but it would be consistent to add such terms. In fact, Dirac showed that the quantization of electric charge could be due to the existence of magnetic monopoles (I can't post another link, so google "magnetic monopole dirac quantization condition"). Maxwell's equations do not tell us whether magnetic monopoles exist or could exist, but the quantization of electric charge could be evidence that magnetic monopoles exist somewhere in the universe.

• In my answer I linked a paper by Hans de Vries where he did as user1247 is saying, and you can check his validity. In motionmoutain ch 18 - Motion in GR, we find also that GravitoElectric is a fundamental field and the GravitoMagnetic is a relativistic effect by the same reason. The motion induces it. To me is a force and not a 'field' (there is no Coriolis field, but force). How the particle can do this? "multiple particles with different velocities, you need to keep track of the electric field in the instantaneous rest frame of each particle" and 'in advance'? – Helder Velez Mar 8 '11 at 3:39

You cannot. B is not just a relativistic side-effect of E. Jackson, Electrodynamics, Section 12.2 has a nice discussion, in which he refutes the "proofs" given in some undergraduate texts.

"The confusion arises chiefly because the Lorentz transformation properties of the force are such that a magnetic-like force term appears when the force in one inertial frame is expressed in terms of the force in another frame. It is tempting to give this extra force term an independent existence and so identify the magnetic field as a separate entity. But such a step is unwarranted without additional assumptions."

Jackson goes on to exhibit an explicit counterexample, based on a Lorentz scalar potential. This field looks like electrostatics (or even Newtonian gravitation!) in the non-relativistic limit. It also has "an apparent magnetic-like force. But there is no independent entity B." So in this "theory" B is indeed only a relativistic effect, but this theory does not apply to Nature.

• 12.2? Kinematics of Decay Products of an Unstable Particle? Or am I reading wrong Electrodynamics by wrong Jackson? – Ruslan Aug 24 '15 at 19:20
• @Ruslan, my ref is Jackson, Classical Electrodynamics, 2nd ed. Section 12.2 heading is "On the Question of Obtaining the Magnetic Field, Magnetic Force, and the Maxwell Equations from Coulomb's law and Special Relativity." – Art Brown Aug 24 '15 at 19:34
• Chapter 12 title is "Dynamics of Relativistic Particles and Electromagnetic Fields". – Art Brown Aug 24 '15 at 19:41
• Hmm, was trying 1st edition. But this isn't present in 3rd too. And I failed to find 2nd. Anyway, there's a followup paper by Kobe, which asserts that the missing assumption to derive Maxwell's equations from Coulomb's law is that charge is a conserved scalar. Anyway, I'd still like to read that Jackson's counterexample you cite, so if you have any link to downloadable 2nd edition, I'd be very thankful if you posted it. – Ruslan Aug 25 '15 at 5:06
• This answer is helpful, but it's an inaccurate summary of what Jackson says. I looked up the relevant passage from the 2nd edition. He does not say that all such proofs are wrong, only that some care is required in spelling out some hidden assumptions. He gives several references to treatments that he says do properly spell out the assumptions: D. H . Frisch and L. Wilets, Am. J. Phys. 24, 574 (1956). J. R. Tessman, Am. J. Phys. 34, 1048 (1966). M. Schwartz, Principles of Electrodynamics, McGraw-Hill, New York (1972), ch. 3. – user4552 Dec 3 '16 at 23:37

With Coulomb's law and special relativity you can derive Ampere's law, which gives you magnetostatics. What's missing for electrodynamics is the displacement current ($\frac{1}{c^2} \frac{\partial E}{\partial t}$), which is a source of magnetic field arising from time-varying electric field, and not a result of the motion of electric charge.

Relativity has only two postulates:

1. The laws of physics are the same in all inertial reference frames
2. All inertial observers measure the same speed for light in vacuum.

Relativity, by itself, does not mandate that electric fields (or electric potential for that matter) must travel at the speed of light. To derive the Maxwell equations, you need an additional postulate, and that is provided by the wave equation (for electric potential) in Section 4 of the reference in Helder's answer. Without this additional postulate (that changes in electric potential propagate at the speed of light), you cannot derive the displacement current from Coulomb's law and relativity alone.

• but Light IS electromagnetic field, and vice-versa – Helder Velez Sep 16 '13 at 14:27
• @Helder Well, if you assume special relativity to derive Maxwell's equations in the way discussed here, you obviously can't refer to the speed of light initially because what light is is not defined until you're done with your derivation. What you can do, though, and in fact must do, is assume that there is a velocity $v$ that appears to be the same in all reference frames. (This is basically what distinguishes Galilean mechanics from SR.) That is, you formulate SR using this velocity $v$ (instead of the speed of light $c$) and derive Maxwell's equations as outlined in... – balu Jun 3 '18 at 16:39
• ...the other posts here. But at some point you will have to make the assumption that $v$ is precisely the velocity at which perturbations of the electromagnetic field propagate. So JxB is absolutely right. – balu Jun 3 '18 at 16:40

by Hans de Vries (*):

The simplest, and the full derivation of Magnetism as a Relativistic side efect of ElectroStatics

He uses only Electrostactic field and the non-simultaneity to obtain the Magnetic Field. He does explain it better than Purcell.

Magnetic field is a side effect of movement in the electric field.

(*) Hans de Vries has a very interesting online book (not yet finished) in his site, and he offers another pearl, not related to this post, but I feel compelled to share: The Lorentz contraction is a real effect and not only 'a referential effect' as we are tempted to believe.

• @JxB I can not comment your answer and quoting "To derive the Maxwell equations, you need an additional postulate, and that is provided by the wave equation (for electric potential) in Section 4 of the reference in Helder's answer. Without this additional postulate (that changes in electric potential propagate at the speed of light), you cannot derive the displacement current from Coulomb's law and relativity alone." – Helder Velez Feb 8 '11 at 15:50
• @JxB continuing previous comment (trouble with Enter key versus ShiftEnter, sorry) electric field = light One can not dissociate electric field from light. "at c speed " here en.wikipedia.org/wiki/Electric_field electr 28 times here: en.wikipedia.org/wiki/Photon_polarization explore the Radiation2D.exe from here www-xfel.spring8.or.jp I do not have any doubt that electric field,and gravity,propagates at c speed. – Helder Velez Feb 8 '11 at 16:15
• @JxB quoting "arising from time-varying electric field, and not a result of the motion of electric charge." Motion is a relative concept and a 'time-varying' electric field is always resultant of charges in motion. – Helder Velez Feb 8 '11 at 16:35
• to keep references togheter: description of Radiation2D.exe method: "NEW MATHEMATICAL METHOD FOR RADIATION FIELD OF MOVING CHARGE" by T. Shintake accelconf.web.cern.ch/accelconf/e02/PAPERS/WEPRI038.pdf – Helder Velez Feb 8 '11 at 18:10

No, you can't. For several reasons. First, if you have E, to get the B field, you need additional assumptions about the structure of the theory, ie in more detail the field strength tensor, see above reply by Lubos. But in addition to this, even if you had the solution for a point charge, to get Maxwell's equations you need to know more than just having one solution. For example that they're linear, second order, and what the symmetry group is. And if you've added that, you can derive the Maxwell equations from these assumptions anyway without even starting with the Coulomb field.

• I have to agree that some extra assumptions are needed. – Philip Gibbs - inactive Jan 23 '11 at 14:45
• +1 for both yours and PhysiXxx's answer: Lubos and all the rest are of course quite right, but it of course depends on what you mean by derive. With the postulates of the OP treated as a formal axiom system: of course you can't; from a physicist's standpoint, where you assume other "reasonable" stuff like linearity, of course you can. – Selene Routley Oct 4 '14 at 3:43

Yes. See Principles of Electrodynamics by Melvin Schwartz. He derives all electrodynamics including Maxwell's equations from Coulomb's Law and Special Relativity.

• He almost made it, but at a point where it is hard to bridge the gap he assorted to God. On page 127, he said, "Appreciating this (antisymmetric tensor is economical), God naturally chose the antisymmetric tensor as His medium of expression." Note that I love this book. It is very concise. I studied it after Griffith's. – verdelite Feb 29 at 5:53

The answer of Luboš Motl is of some help, in that it shows how to bring in the sorts of insights which relativity offers, but nevertheless it opens with its overall conclusion, and that conclusion is wrong. It is wrong largely for the reasons briefly indicated in the answer of WIMP.

The question is important, and it is important to get the answer right. The question is:

Can Maxwell's equations be derived using only Coulomb's Law and Special Relativity?

The answer is: no, because plenty of other field theories that respect Special Relativity can be invented, such that they reproduce Coulomb's Law in the inertial frame of a given point charge.

However, what one can say, is that classical electromagnetism (i.e. Maxwell's equation and the Lorentz force equation, or any formulation equivalent to this, such as a Lagrangian formulation) is among the simplest field theories that respect Special Relativity and include Coulomb's law. The definition of 'simplest' here is admittedly imprecise.

The main reason why you can't derive Maxwell from 'Coulomb + S.R.' is that you would not know whether to include acceleration effects in the relationship between potentials and charges.

Now I will 'lift the lid' a little on the theoretical physics here. A very good (not the only) mathematical way to ensure that any piece of physics respects Special Relativity (S.R.) is to restrict oneself to tensorial expressions in everything you propose and write down. Here 'tensorial' includes tensors of rank zero, i.e. scalars, but not just any old scalars: they would be Lorentz-invariant scalars. It also includes 4-vectors and second- and higher- rank tensors. When taking derivatives, you use the covariant gradient operator $$\partial_a$$, and then you have a tool kit for constructing differential equations that respect S.R.

So the 'simplest' field theory might be one such that particles can have a Lorentz-invariant scalar property called charge $$q$$, and the force on a charged particle is independent of the 4-velocity $$u^a$$ of the particle. The trouble is that you quickly find that in such a theory the force on a particle cannot change the velocity of a particle without also changing its mass. Exploring further, you try allowing the 4-force $$f^a$$ to be dependent on the 4-velocity through a simple linear equation involving a scalar field $$\phi$$, such as $$f^a = q \phi u^q$$ (?). Still no good (mass changes again). So you are led to try a second-rank tensor $$F^{ab}$$ for the field, because it is the simplest thing, other than a scalar, which can take a 4-vector $$u^a$$ as input and give back a 4-vector force:

$$f^a = q F^{a\mu} u_\mu$$

Now it's ok: the force is mass-preserving as long as $$F^{ab}$$ is antisymmetric. Good! An antisymmetric tensor is the simplest type of second-rank tensor. Next we want a differential equation for this field: try the simplest thing, which is to take the divergence, and you are well on the way to Maxwell's equations. If we now bring in Coulomb's law (and this is where it comes in), then you are guaranteed to get two of Maxwell's equations if you restrict the source term in your differential equation to only a single term proportional to charge density and 4-velocity. Coulomb's law does not itself tell you not to add in further terms to do with 4-acceleration.

By this approach we do not arrive inexorably at Maxwell's equations, but one does find that they are arguably the simplest that include the property of charge conservation and that allow a mass-preserving force (in technical language, a pure force).

Among other field theories that one encounters there is one that is much like Maxwell but includes magnetic monopoles. This arises very naturally, in the theoretical treatment, and is certainly a serious candidate possibility for how the physical world really works. It is somewhat less simple in that one loses the nice property of writing the field tensor as a 4-curl of a 4-vector field (the 4-potential), and the theory no longer respects symmetry under space inversion (parity). See Jackson's book on electromagnetism for a discussion. If there are in fact magnetic monopoles, as many versions of quantum field theory suggest, then the puzzle is why electric monopoles are so much more abundant than magnetic monopoles.

However, I would like to underline that this magnetic monopole issue is far from the only reason Maxwell's equations are not fully derivable from Coulomb's law and S.R. The other reasons include that one can easily imagine that the field equations involve higher-order derivatives of the motion of the particle; S.R. on its own cannot tell you that they don't. By starting out with a Lagrangian approach, one can introduce further constraints, such as invariance leading to conservation laws, and then electromagnetism is quite tightly, but still not fully, constrained. Fundamentally, what S.R. can tell you is that a field which provides a force independent of a body's velocity cannot be the whole story about the physics. Such a field (such as the electric field) must be in partnership with further effects which do depend on a body's velocity.

There are few articles that show that conservation/continuity equation for the electric charge is sufficient to derive the whole set of Maxwell equations. See this ref and quotes for example; https://pdfs.semanticscholar.org/3251/31eadb62c8fdfdaaad7b21a308992ff3a4d2.pdf ''How to obtain the covariant form of Maxwell’s equations from the continuity equation.... Therefore, a circular process seems to be unavoidable in electromagnetism: ρ and J imply E and B which, in turn, imply new ρ1 and J1, and so on. Because of this circular characteristic, it is not clear if E and B (satisfying Maxwell’s equations) are a consequence of ρ and J (satisfying the continuity equation) or vice versa. According to the referee it seems a matter of taste to say which one is a consequence of the other. In other words: from referee’s comment we could conclude that the connection between sources and fields is a little bit like the egg and hen problem: who was first?''.

Also starting from Coulomb law for static electricity and taking into account the fact that action can't travel faster than light, the use of the retarded integral produces the complete set of Maxwell equations. https://en.wikipedia.org/wiki/Li%C3%A9nard%E2%80%93Wiechert_potential.

Thus, retardation itself gives rise to a force that is normal to the motion(velocity), proportional to it, and decaying as the inverse square of the distance- that is the magnetic field by definition. It also gives rise to a two component force- an electric and magnetic fields proportional to acceleration, that decay only as the inverse of the distance(not the inverse squared) and this is radiation by definition. One can therefore deduce that magnetism and radiation are emergent phenomenon caused by the finiteness of the speed of propagation of the forces involved.

Some answers pointed to connection with Gravito-magnetism and relativity. I think this comes from the fact that Newton's law of gravitation can be treated in a similar way to Coulomb law, giving rise to a set of equations similar to Maxwell equations. These are the gravito-magnetic equations and are in fact also derivable from general relativity for weak fields. https://en.wikipedia.org/wiki/Gravitoelectromagnetism

As I understand your idea, you are asking if it is possible to recover all Maxwell's equations only using Lorentz transformations and using the existence of electric field. The answer is no. An heuristic example is this: If you have a circular unidimensional wire with a variable current $$I(t)$$, there is no Lorentz transformation to produce the magnetic field of this system starting only from a electric field, because the electric charge of the circular wire is moving in an non inertial way.

• I think otherwise. For each of the infinitesimal segment of the wire, for a short infinitesimal time interval delta t, in the co-moving inertial frame whose speed is the same as the electrons in the wire at that instance, we can talk about Lorentz transformation leading to Maxwell equations. Also the 4-potential only depends on the instant current only and not on its derivatives thus we can do that. – verdelite Mar 11 '19 at 16:16